In answer to Pete L. Clark's question Must a ring which admits a Euclidean quadratic form be Euclidean? on Euclidean quadratic forms, I gave an example in seven variables, repeated below. Pete's Euclidean property is simply that for any point $\vec x \in \mathbf Q^7$ but $\vec x \notin \mathbf Z^7,$ we require that there be at least one $\vec y \in \mathbf Z^7$ such that $$ q(\vec x - \vec y) < 1. $$

[Edit: This is the definition for positive definite integral quadratic forms. --PLC]

I think my answer works (and the easier 6 variable one), based on extensive computer calculations, and Pete has been too polite to express much doubt.

EDIT. I think it wise to describe what I am completely certain about and what is unclear. What I did is make a cubic grid, where each variable takes on values $\frac{i}{M}$ for $0 \leq i < M.$ So that makes a grid with $M^7$ points. For each point $\vec r$ in the grid, I find the 128 different values of $q(\vec r - \vec y)$ for
$\vec y \in \mathbf Z^7$ and all coordinates of $\vec y$ are either 0 or 1. For that point $\vec r,$ I take the smallest of the 128 values. Now, for every $M$ I have tried, and for every $\vec r$ in the grid, this best value out of 128 has never been larger than $\frac{7}{8}.$ Now, using the fact that for any $\vec x \in \mathbf Q^7$ that is not in the grid, there is some point $\vec r$ such that $ | \vec r -\vec x | \leq \frac{1}{2 M \sqrt 7},$
I get that I can always find a $\vec y \in \mathbf Z^7$ such that
$$ g(x-y) \leq \frac{7}{8} + \frac{1}{7 M} + \frac{1}{14 M^2}, $$ using Cauchy-Schwarz and the maximum eigenvalue of $Q$ being 2. Anyway, this does not show that the Euclidean minimum is really $\frac{7}{8},$ although I believe it is. What it does show is that the Euclidean minimum is less than 1, as soon as $M \geq 2.$

2 Answers
2

Consider the form
$$ Q(x) = 2q(x) = (x_1+x_2)^2 + (x_2+x_3)^2 + \ldots + (x_7+x_1)^2.$$
You have to show that it has Euclidean minimum $\frac74$ attained at $X_1 = x_1+x_2 = \frac12$, ..., $X_7 = x_7 + x_1 = \frac12$, but unfortunately not over the lattice ${\mathbb Z}^7$, where it would be trivial, but over a lattice of index $2$ defined by the condition $y_1 + y_2 + \ldots + y_7 \equiv 0 \bmod 2$. It is clear that $(\frac12, \ldots, \frac12)$ has Euclidean minimum $\frac 74$, and it remains to show that all other points in the fundamental domain of the lattice have a minimum at most $\frac74$.

This is not difficult to see: assume you have the point $(\frac12 + \delta, \frac12 + \varepsilon, ...)$ for small $\delta, \varepsilon \ge 0$. Subtracting the point $(1,1,0,\ldots, 0)$ you will get a point with coordinates $(-\frac12 + \delta, -\frac12 + \varepsilon, ...)$. Repeating this procedure you will eventually reach a point with
$|X_2|, \ldots, |X_7| \le \frac12$. If $|X_1| \le\frac12$, we are done. If not, you can
make $|X_1| < \frac12$ at the cost of making another coordinate $> \frac12$ in absolute values. If you think about this for a minute you will see that we can always reach a point of the form
$$ X = \Big(\frac12 + \delta_1, \frac12 - \delta_2, . . . , \frac12 - \delta_7\Big)$$
with $0 \le \delta_j \le \frac12$ and $\delta_1 \le \delta_i$ for all $i > 1$. It remains to show that $q(X) \le \frac74$, which is equivalent to
$$ \delta_1 - \delta_2 - \ldots - \delta_7 + \delta_1^2 + \ldots + \delta_7^2 \le 0. $$
Does this inequality hold?